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Alex777 [14]
3 years ago
5

Can someone help me ?

Mathematics
1 answer:
TEA [102]3 years ago
3 0
We use the <span>Pythagorean Theorem method.

</span><span><span>a^2 </span>+ <span>b^2 </span>= <span>c^2
</span></span>
9^2 + b^2 = 15^2

9^2 + 12^2 = 15^2

The length of side B is 12 cm
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Help this is due tomorrow
nlexa [21]

Answer:

n = 1.40 (second option)

8 0
3 years ago
Read 2 more answers
IAS question - ❓❓❓❓❓❓A man starts frm his house with some money to visit four temples.1. As soon as he enters temple 1 his money
Natali5045456 [20]

Answer:

Rs 93.75

Step-by-step explanation:

Solution:

A man starts from his house with some money to visit four temples.

<u>1. As soon as he enters temple 1 his money gets doubled. </u>

Let he had money at first = x

After enter the 1st temple his money = 2x

He donated = Rs 100

Money left after donation at 1st temple = 2x-100

<u>2. At temple 2 again his money is doubled and he donates Rs 100</u>

He had at 2nd temple = 2(2x-100) =4x-200

Donated = Rs 100

Money left after donation at 2nd temple = 4x-200-100=4x-300

<u>3. His money is doubled again at temple 3 and he donates Rs 100.</u>

He had at 3rd temple = 2(4x-300)=8x-600

Donated = Rs 100

Money left after donation at 3rd temple = 8x-600-100=8x-700

<u>4. At temple 4 again his money gets doubled. He donates 100 rs. again and returns home empty handed.</u>

He had at 4th temple = 2(8x-700)= 16x-1400

Donated = Rs 100

Money left after donation at 4th temple = 16x-1400-100 = 16x-1500

<u>As he returned empty hand or with zero money, equation will be:-</u>

16x-1500=0\\ \\ Adding\ both\ sides\ by\ 1500\\ \\ 16x=1500\\ \\ Dividing\ both\ sides\ by\ 16\\ \\ x=Rs \ 93.75

Thus, he had Rs 93.75 when he started from his house.

6 0
3 years ago
If the word is a proper noun, choose the answer that begins with a capital letter. If the word is a common noun, choose
ad-work [718]
School is not a a proper noun so it will start with a lowercase letter
4 0
3 years ago
20 POINTS!!! <br><br> belongs to a: <br> pyramid<br> cube<br> rectangular prism<br> triangular prism
Klio2033 [76]

The middle row is long on one side and short in the other side which is the shape of a rectangle.

The net is a rectangular prism

8 0
3 years ago
Can someone explain this to me please
IrinaVladis [17]

Answer:

c. 36·x

Step-by-step explanation:

Part A

The details of the circle are;

The area of the circle, A = 12·π cm²

The diameter of the circle, d = \overline {AB}

Given that \overline {AB} is the diameter of the circle, we have;

The length of the arc AB = Half the the length of the circumference of the circle

Therefore, we have;

A = 12·π = π·d²/4 = π·\overline {AB}²/4

Therefore;

12 = \overline {AB}²/4

4 × 12 = \overline {AB}²

\overline {AB}² = 48

\overline {AB} = √48 = 4·√3

\overline {AB} = 4·√3

The circumference of the circle, C = π·d = π·\overline {AB}

Arc AB = Half the the length of the circumference of the circle = C/2

Arc AB = C/2 = π·\overline {AB}/2

\overline {AB} = 4·√3

∴ C/2 = π·4·√3/2 = 2·√3·π

The length of arc AB = 2·√3·π cm

Part B

The given parameters are;

The length of \overline {OF} = The length of \overline {FB}

Angle D = angle B

The radius of the circle = 6·x

The measure of arc EF = 60°

The required information = The perimeter of triangle DOB

We have;

Given that the base angles of the triangles DOB are equal, we have that ΔDOB is an isosceles triangle, therefore;

The length of \overline {OD} = The length of \overline {OB}

The length of \overline {OB} = \overline {OF} + \overline {FB} = \overline {OF} + \overline {OF} = 2 × \overline {OF}

∴ The length of \overline {OD} = 2 × \overline {OF} = The length of \overline {OB}

Given that arc EF = 60°, and the point 'O' is the center of the circle, we have;

∠EOF = The measure of arc EF = 60° = ∠DOB

Therefore, in ΔDOB, we have;

∠D + ∠B = 180° - ∠DOB = 180° - 60° = 120°

∵ ∠D = ∠B, we have;

∠D + ∠B = ∠D + ∠D = 2 × ∠D = 120°

∠D = ∠B = 120°/2 = 60°

All three interior angles of ΔDOB = 60°

∴ ΔDOB is an equilateral triangle and all sides of ΔDOB are equal

Therefore;

The length of \overline {OD} = The length of \overline {OB} = The length of \overline {DB}  = 2 × \overline {OF}

The perimeter of ΔDOB = The length of \overline {OD} + The length of \overline {OB} + The length of \overline {DB} = 2 × \overline {OF} + 2 × \overline {OF} + 2 × \overline {OF} = 6 × \overline {OF}

∴ The perimeter of ΔDOB = 6 × \overline {OF}

The radius of the circle = \overline {OF} = 6·x

∴ The perimeter of ΔDOB = 6 × 6·x = 36·x

3 0
3 years ago
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